## Algebra 2 (1st Edition)

Since the ordering doesn't matter in the selection, we use combinations. $_nP_r=\frac{n!}{(n-r)!}$, hence here: $_{12}P_2=\frac{12!}{(12-2)!}=\frac{12!}{10!}=12\cdot11=132$ We know that $_nC_r=\frac{n!}{r!(n-r)!}$. Hence $_{280}C_5=\frac{280!}{5!275!}=13836130056$